#2283

Easy

LeetCode Problem Workspace

Check if Number Has Equal Digit Count and Digit Value

Determine if a number's digits correspond to their counts in the string representation.

Problem Statement

You are given a string num of length n, consisting of digits. You need to determine if, for each index i in the range 0 <= i < n, the value of num[i] corresponds to the number of times the digit i appears in num. Return true if this condition holds for every index, otherwise return false.

For example, for num = "1210", the digit 0 occurs once, digit 1 occurs twice, digit 2 occurs once, and digit 3 occurs zero times. This satisfies the condition, so the result is true. On the other hand, num = "030" doesn't satisfy the condition because the frequency of digits does not match the required counts.

Examples

Example 1

Input: num = "1210"

Output: true

num[0] = '1'. The digit 0 occurs once in num. num[1] = '2'. The digit 1 occurs twice in num. num[2] = '1'. The digit 2 occurs once in num. num[3] = '0'. The digit 3 occurs zero times in num. The condition holds true for every index in "1210", so return true.

Example 2

Input: num = "030"

Output: false

num[0] = '0'. The digit 0 should occur zero times, but actually occurs twice in num. num[1] = '3'. The digit 1 should occur three times, but actually occurs zero times in num. num[2] = '0'. The digit 2 occurs zero times in num. The indices 0 and 1 both violate the condition, so return false.

Constraints

n == num.length
1 <= n <= 10
num consists of digits.

Solution Approach

Count Frequencies Using a Hash Table

The problem is naturally a hash table problem. First, create a frequency map of the digits in the string. Then, for each index, check if the digit's value matches its frequency in the map. This is efficient because the hash table provides constant-time lookups.

Handle Edge Cases

Pay attention to edge cases like single-digit numbers or numbers with many repeated digits. These cases often require careful counting, as one missed count can lead to an incorrect result.

Efficiency Considerations

Given the constraints, with n <= 10, even a brute force solution will be efficient. However, it’s important to understand that this problem can be optimized using hash tables to store and compare counts for an overall better performance.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) due to the need to iterate over the string and create the frequency map. Space complexity is O(n) for the frequency map, where n is the length of the string num.

What Interviewers Usually Probe

Look for understanding of hash table usage to track digit frequencies.
Check if the candidate can identify potential edge cases and handle them correctly.
Verify the candidate's ability to optimize solutions by recognizing that this problem can be solved in linear time.

Common Pitfalls or Variants

Common pitfalls

Forgetting to correctly count the digits in num can lead to wrong answers.
Confusing the digit's value with its index in the string.
Not handling edge cases where digits might repeat or be absent.

Follow-up variants

Consider larger numbers where the string length exceeds 10.
Explore how to handle larger digit ranges with more efficient counting techniques.
What would happen if the input was allowed to be an arbitrary large number, and how would the approach change?

FAQ

What is the main technique used in solving the "Check if Number Has Equal Digit Count and Digit Value" problem?

The main technique is using a hash table to count the frequency of digits in the string and then comparing the count with the value of each digit at the respective index.

How do you handle edge cases in this problem?

Edge cases, like strings with repeated digits or a single character, need to be handled by ensuring the correct counting of digits and comparing each index’s value accurately.

Can this problem be solved with brute force?

Yes, but using a hash table to store digit frequencies will give a more efficient solution with linear time complexity.

Why does the solution work for the problem "Check if Number Has Equal Digit Count and Digit Value"?

The solution works by creating a frequency map of digits, allowing quick lookups and comparisons between each index’s digit value and its count in the string.

What other algorithms or data structures could be used for this problem?

Aside from hash tables, one could use arrays or counters to count occurrences, but hash tables offer the most optimal and intuitive solution.

terminal

Solution

Solution 1: Counting + Enumeration

We can use an array $\textit{cnt}$ of length $10$ to count the occurrences of each digit in the string $\textit{num}$. Then, we enumerate each digit in the string $\textit{num}$ and check if its occurrence count equals the digit itself. If this condition is satisfied for all digits, we return $\text{true}$; otherwise, we return $\text{false}$.

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class Solution:
    def digitCount(self, num: str) -> bool:
        cnt = Counter(int(x) for x in num)
        return all(cnt[i] == int(x) for i, x in enumerate(num))

Continue Practicing

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